This note primarily delves into incompressible flow momentum equation and Surface of water waves and boundary condition and Variational Formulation.

The Equations for Water Waves

incompressible flow momentum equation:

\[ \begin{gathered} \nabla \cdot \mathbf{u}=0 \\ \frac{D \mathbf{u}}{D t}=\frac{\partial \mathbf{u}}{\partial t}+(\mathbf{u} \cdot \nabla) \mathbf{u}=-\frac{1}{\rho} \nabla p-g \mathbf{j} \end{gathered} \] velocity potential: \(\varphi\) with \(\mathbf{u}=\nabla \varphi\). vorticity: \(\boldsymbol{\omega}= \text{curl } \mathbf{u}\)

\(\nabla \times \text{ or } \text{curl}\) \[ \frac{\partial \omega}{\partial t}+\nabla \times(\omega \times \mathbf{u})=0 \] we start form \[ \frac{D \vec{\omega}}{D t}=\partial_t \vec{\omega}+(\vec{u} \cdot \nabla) \vec{\omega} \tag{1} \] \[ \partial_t \vec{\omega}+\nabla \times(\vec{\omega} \times \vec{u})=0\tag{2} \] from (1) (2) we have \[ \frac{D \vec{\omega}}{D t}=(\vec{u} \cdot \nabla) \vec{\omega}-\nabla \times(\vec{\omega} \times \vec{u})\tag{3} \] the first term of (3) \[ \begin{aligned} (\vec{u} \cdot \nabla) \vec{\omega} & =\left(u_i \vec{e}_i \cdot \partial_j \vec{e}_j\right) \omega_k \vec{e}_k \\ & =u_i \partial_i \omega_j \vec{e}_j \end{aligned} \] the second term of \((3)\) \[ \begin{aligned} -\nabla \times(\vec{\omega} \times \vec{u}) & =-\partial_i \vec{e}_i \times\left(\omega_j \vec{e}_j \times u_k \vec{e}_k\right) \\ & =-\partial_i \vec{e}_i \times\left(\omega_j u_k \varepsilon_{j k l} \vec{e}_l\right) \\ & =-\partial_i\left(\omega_j u_k\right) \varepsilon_{j k l} \varepsilon_{i l m} \vec{e}_m \\ & =\partial_i\left(\omega_j u_k\right) \varepsilon_{j k l} \varepsilon_{l i m} \vec{e}_m \\ & =\partial_i\left(\omega_j u_k\right)\left(\delta_{j i} \delta_{k m}-\delta_{j m} \delta_{k i}\right) \vec{e}_m \end{aligned}\tag{4} \] divide (4) into two parts \[ \underbrace{\partial_i\left(\omega_j u_k\right) \delta_{j i} \delta_{k m} . \vec{e}_m}_{i=j\quad t=m} \quad \text { and } \underbrace{-\partial_i\left(\omega_j u_k\right) \delta_{j m} \delta_{k i} \vec{e}_m}_{j=m\quad k=i} \] then \[ \underbrace{\partial_i\left(\omega_i u_k\right) \vec{e}_k}_{(4)-1} \text { and }-\underbrace{-\partial_i\left(\omega_j u_i\right) \vec{e}_j}_{(4)-2} \] (4)-1 can be written as \[ \begin{align*} \begin{array}{ll|} &\omega_i \partial_i u_k \vec{e}_k+u_k \partial_i \omega_i \vec{e}_k\\ &u_k \vec{e}_k \partial_i \omega_i\\&=\vec{u} \cdot(\nabla \cdot \vec{\omega})=0\\ \end{array} \begin{array}{ll} &\text{notes:} \\ & \nabla \cdot(\nabla \times \vec{a})=0 \\ & \nabla \times(\nabla \varphi)=0 \\ &\vec{\omega}=\nabla \times \vec{u}\\ & \nabla \cdot \vec{u}=0 \\ & \nabla \cdot \overrightarrow{\omega}=0 \end{array} \end{align*} \] (4)-1 becomes \[ \omega_i d_i u_k \vec{e}_k \] and (4)-2 \[ \begin{aligned} & -\left(\omega_j \partial_i u_i \vec{e}_j+u_i \partial_i \omega_j \vec{e}_j\right) \\ & \omega_j \overrightarrow{e_j} \partial_i u_i=\vec{\omega} \cdot(\nabla \cdot \vec{u})=0 \end{aligned} \] (4)-2 becomes \[ -u_i \partial_i \omega_j \vec{e}_j \] Therefore equation (3) becomes \[ \begin{aligned} & u_i \partial_i \omega_j \vec{e}_j+\omega_i \partial_i u_k \vec{e}_k-u_i \partial_i \omega_j \vec{e}_j \\ & =\omega_i \partial_i \vec{u}_k \vec{e}_k \\ & =(\vec{w} \cdot \nabla) \vec{u} \end{aligned} \] finally \[ \frac{D \omega}{D t}=(\vec{\omega} \cdot \nabla) \vec{u} \]

Surface of water waves

interface: \(f\left(x_1, x_2, y, t\right)=0\). normal velocity of a surface \[ \frac{-f_t}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}} . \] The normal velocity of the fluid is \[ \frac{u_1 f_{x_1}+u_2 f_{x_2}+v f_y}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}} . \]


\[\begin{aligned}& \frac{D f}{D t}=f_t+u_1 f_{x_1}+u_2 f_{x_2}+v f_y=0 \\& f_t=-u_1 f_{x_1}-u_2 f_{x_2}-v f_y \\& \frac{-f_t}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}}=\vec{u} \cdot \vec{n}_s\end{aligned}\] \(\vec{n}_s\) the normal vector of surfacewaterwaves-1.png \(\frac{-f_t}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}} \rightarrow\) we take \(f\) as function of surface here it's crucial to emphasize that f is only taken as a scalar function which is represent the surface. therefore \(\frac{\partial f}{\partial t}\) is velocity of surface (also can say it is the normal vector of surface but not the unit vector)

unit vector \(-\frac{f_t}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}}\) and the \(-\) sign represent the direction toward the exterior surface. \[\frac{u_1 f_{x_1}+u_2 f_{x_2}+v f_y}{\sqrt{f_{x_1}^2+f_{x_2}^2+f_y^2}}\] which is represent the velocity of fluid (scalar) along the normal direction of the surface

This shows that particles in the surface remain there.

Explanation : particles in the surface are always in the surface

others description \(y=\eta\left(x_1, x_2, t\right)\): \[ f\left(x_1, x_2, y, t\right) \equiv \eta\left(x_1, x_2, t\right)-y \tag{4} \]


Explanation for equation (4): in \(y=\eta\left(x_1, x_2, t\right)\), there are two surface (interface, fluid surface), which are actually identical, but indicate two different properties. where \(y\) is the interface(it's independent of fluid) and the other \(\eta\) is the fluid surface.

boundary condition

Assumption: 1. surface tension is neglected 2. free surface

on free surface

\[ \left.\begin{array}{r} \eta_t+\varphi_{x_1} \eta_{x_1}+\varphi_{x_2} \eta_{x_2}=\varphi_y, \\ \varphi_t+\frac{1}{2}\left(\varphi_{x_1}^2+\varphi_{x_2}^2+\varphi_y^2\right)+g \eta=0 . \end{array}\right\} \text { on } y=\eta\left(x_1, x_2, t\right) \text {. } \] where \[ \begin{aligned} & \varphi_{x_1}, \varphi_{x_2}, \varphi_y=\left(u_1, u_2, v\right) \\ & \varphi_y=v=\frac{d y}{d t} \end{aligned} \]

at bottom

Let the normal vector of surface \(\vec{n}\), the projection of velocity onto normal vector is zero: \[ \begin{gathered} \vec{u} \cdot \vec{n}=0 \quad\left(\text { or } u_n=0\right) \\ \vec{u}=\nabla \varphi \quad \vec{n}=\left(\frac{\partial y}{\partial y}, \frac{\partial y}{\partial x_1}, \frac{\partial y}{\partial x_2}\right) \quad\left(y=h_0\left(x_1, x_2, t\right)\right) \end{gathered} \] therefore \[ \varphi_y+\varphi_{x_1} h_{0, x_1}+\varphi_{x_2} h_{0, x_2}=0 \] For the case: the flat horizontal bottom \[ \varphi_y=0 \text { on } y=-h_0 . \]

Variational Formulation

the variational principle \[ \begin{gathered} \delta \iint_R L d x d t=0, \\ L=-\rho \int_{-h_0}^\eta\left\{\varphi_t+\frac{1}{2}(\nabla \varphi)^2+g y\right\} d y, \end{gathered} \]

For a small change \(\delta \varphi\) in \(\varphi\), \[ \begin{aligned} -\delta \iint \frac{L}{\rho} d \mathbf{x} d t= & \iint_R\left\{\int_{-h_0}^\eta\left(\delta \varphi_1+\nabla \varphi \cdot \nabla \delta \varphi\right) d y\right\} d \mathbf{x} d t \\ = & \iint_R\left\{\frac{\partial}{\partial t} \int_{-h_0}^\eta \delta \varphi d y+\frac{\partial}{\partial x_i} \int_{-h_0}^\eta \varphi_{x_1} \delta \varphi d y\right\} d \mathbf{x} d t \\ & -\iint_R\left\{\int_{-h_0}^\eta\left(\varphi_{x_1 x_1}+\varphi_{y y}\right) \delta \varphi d y\right\} d \mathbf{x} d t \\ & -\iint_R\left[\left(\eta_t+\varphi_{x_1} \eta_{x_1}-\varphi_y\right) \delta \varphi\right]_{y=\eta} d \mathbf{x} d t \\ & -\iint_R\left[\left(\varphi_{x_1} h_{0 x_1}+\varphi_y\right) \delta \varphi\right]_{y=-h_0} d \mathbf{x} d t . \end{aligned} \] unsolved (some difference between my derivation and these above)


\[\begin{aligned}& \nabla \varphi=\left(\frac{\partial}{\partial x_i} \varphi, \frac{\partial}{\partial y} \varphi\right) \\& \delta\left(\frac{1}{2} \nabla \varphi\right)^2=\nabla \varphi \cdot \nabla \delta \varphi . \\&=\frac{\partial}{\partial x_i} \varphi \cdot \frac{\partial}{\partial x_i} \delta \varphi+\frac{\partial}{\partial y} \varphi \cdot \frac{\partial}{\partial y} \delta \varphi \\&=\underbrace{\varphi_{x_i} \frac{\partial}{\partial x_i} \delta \varphi}_{(5 \cdot 1 \cdot 1)}+\underbrace{\varphi_y \frac{\partial}{\partial y} \delta \varphi}_{(5.1.2)}\end{aligned}\] -> 5.1.1 \[\varphi_{x_i} \frac{\partial}{\partial x_i} \delta \varphi=\frac{\partial}{\partial x_i}\left(\varphi_{x_i} \delta \varphi\right)-\varphi_{x_i x_i} \delta \varphi \quad(5.2 .1)\] ->5.1.2 \[\varphi_y \frac{\partial}{\partial y} \delta \varphi=\frac{\partial}{\partial y}\left(\varphi_y \delta \varphi\right)-\varphi_{y y} \delta \varphi \quad \text { (5.2.2) }\] combine \((5.2.1)(5.2.2)\) we have \[\frac{\partial}{\partial x_i}\left(\varphi_{x_i} \delta \varphi\right)-\left[\varphi_{x_i x_i}+\varphi_{y y}\right] \delta \varphi+\underbrace{\frac{\partial}{\partial y}\left(\varphi_y \delta \varphi\right)}_{5.3.1}(5.3)\] ->5.3.1 apply integral with respect to \(y\) to 5.3.1 \[\begin{aligned}& \int \frac{\partial}{\partial y}\left(\varphi_y \delta \varphi\right) d y \\& \quad=\left[\varphi_y \delta \varphi\right]_{y=-h_0}^{y=\eta} \\& \quad=\left[\varphi_y \delta \varphi\right]_{y=\eta}-\left[\varphi_y \delta \varphi\right]_{y=-h_0}\end{aligned}\]

\[ \begin{aligned} & \varphi_{x_i x_i}+\varphi_{y y}=0, \quad-h_0<y<\eta, \\ & \eta_t+\varphi_{x_i} \eta_{x_i}-\varphi_y=0, \quad y=\eta, \\ & \varphi_{x_i} h_{0 x_i}+\varphi_y=0, \quad y=-h_0 \end{aligned} \]